#### To determine

Whether the geometric series is convergent or divergent and obtain the sum if the series is convergent.

#### Answer

The series is divergent.

#### Explanation

**Given:**

The series is
∑n=0∞e2n6n−1
.

**Result used:**

The geometric series
∑n=1∞arn−1 (or) a+ar+ar2+⋯
is divergent if
|r|≥1
, where *a* is the first term and *r* is the common ratio of the series.

**Calculation:**

Consider the given series
∑n=0∞e2n6n−1
.

∑n=0∞e2n6n−1=e2⋅161−1+e2⋅262−1+e2⋅363−1+e2⋅464−1+⋯=e260+e461+e662+e863+⋯=e2+e46+e636+e8216+⋯

Here, the first term of the series is
a=e2
and the common ratio of the series is,

r=Second termFirst term=e46e2=e46⋅1e2=e26

The absolute value of *r* is,

|r|=|e26|=e26=1.2315>1

Since
|r|>1
and by using Result (2) stated above, the series is divergent.

Therefore, the series is divergent.